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EXPLANATION OF KRYPTON IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) May 27 , 2015 Krypton is a chemical element with symbol Kr and atomic number 36. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this image of Krypton including the following ground state electron configuration: 1s22s22p63s23p63d104s24p6. According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of krypton (from (E1 to E26) are the following: E1 = 13.99, E2 = 24.36, E3 = 36.95 , E4 = 52.5 , E5 = 64.7, E6 = 78.5, E7 = 111, E8= 125.8, E9 = 230.85, E10 = 268.2, E11 = 308 , E12 = 350, E13 = 391, E14 = 447 , E15 = 492, E16 = 541, E17 = 592, E18 = 641, E19 = 786, E20 = 833 , E21 = 884 E22 = 937, E23 = 998 , E24 = 1051, E25 = 1151, E26 = 1205.3 Firstly we examine the -( E1 + E2 + Ε3 + Ε4 + Ε5 + Ε6 ) = -271 = E(4p6) Here the E(4p6) represents the binding energy of the six outermost electrons. Then, we observe that -( Ε7 + E8 ) = - 236.8 = E(4s2) - ( E9 + … + E18 ) = - 4261 = E(3d10). -(E19 +…+ E24) = - 5489 = E(3p6) -(E25 + E26 ) = -2356.3 = E(3s2) It is of interest to note that in the absence of data (from E27 to E36) one can write -( E27 +…+ E32 ) = E(2p6) -(E33 + E34) = E(2s2) -(E35 + E36) = E(1s2) Such situations are analogous to those of copper. ( See my EXPLANATION OF COPPER IONIZATIONS ). For understanding better the ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. ' ' EXPLANATION OF - ( E1 +…+ E6 ) = -271 = E(4p6) Here the binding energy E(4p6) of the six outermost electrons is given by applying my formula of 2008. The charges (-30e) of the electrons (1s22s22p63s23p63d104s2) screen the nuclear charge (+36e) and for a perfect screening we would have an effective ζ = 6. However the electrons 4p6 of the three orbitals repel the 4s2 leading to the deformation of 4s2 . Thus ζ > 6. Under this condition we may write (E1 +… + E6) = 271 eV = -E(4p6) = -327.21 )ζ2 + (16.95) ζ - 4.1 / n2 Since n = 4 the above equation could be written as 5.1 ζ2 – 3.178 ζ - 270.23 = 0 Then solving for ζ we get ζ = 7.6 > 6 ' ' EXPLANATION OF -( Ε7 + E8 ) = - 236.8 = E(4s2) ''' Here the E(4s2) represents the binding energy of the two electrons with opposite spin given by applying my formula of 2008. The charges (-28e) of the inner electrons (1s22s22p63s23p63d10) screen the nuclear charge (+36e) and for a perfect screening we would have ζ = 8. However according to the quantum mechanics the two electrons of 4s2 penetrate the 3d10 and lead to the deformation of 3d10 . Thus we must observe that ζ > 8. Under this condition we may write (Ε7 + E8 ) = 236.8 = -E(4s2) = - + (16.95)ζ - 4.1 / n2 Since n = 4 we may rewrite 1.7ζ2 -1.059ζ -236.54 = 0 Then solving for ζ we get ζ = 12.11 > 8 ' ' '''EXPLANATION OF ( E9 + … + E18 ) = 4261 eV = -E(3d10) Here the E(3d10) represents the binding energy of the 10 electrons (3d10) . Note that the 10 electrons create five orbitals with electrons of opposite spin given by applying my formula of 2008. The charges (-18e) of the inner electrons (1s22s22p63s23p6) screen the nuclear charge (+36e) and for a perfect screening we would have ζ = 18. Under this condition we may write ' '''E9 +..+ E18 ) = 4261 eV = - E(3d10) = -5+ (16.95)ζ - 4.1 / n2 So using n = 3 we may rewrite 15.1167ζ2- 9.4167ζ - 4258.72 = 0 Surprisingly solving for ζ we get ζ = 17.01 < 18 , which cannot exist . In fact, the 10 electrons of the five orbitals make a complete spherical sub- shell leading to a perfect screening with ζ = 18. Thus using ζ = 18 we expect to determine n > 3. Under this condition we may write E9 +..+ E18 ) = 4261 eV = -E(3d10) = -527.21)182 - (16.95)18 + (4.1) / n2 Then solving for n we get n = 3.16 > 3 . ' ' '''EXPLANATION OF ( E19 + … + E24 ) = 5489 eV = -E(3p6)' Here the E(3p6) represents the binding energy of the 6 paired electrons given by applying my formula of 2008. ' '''The charges (-12e) of the twelve inner electrons like (1s22s22p63s2 ) screen the nuclear charge (+36e) and for a perfect screening we would have an effective Zeff = ζ = 24. However the 6 paired electrons ( 3p6 ) repel the 3s2 electrons and lead to the deformation of shells with ζ > 24. Under this condition we may write ( E19 +…+ E24) = 5489 eV' = '-E(3p6) '= '-3+(16.95)ζ - 4.1 / n2 Since n = 3 the above equation can be written as 9.07ζ2 - 5.65ζ - 5487.63 = 0 Then, solving for ζ we get ζ = 24.91 > 24 . ' ' '''EXPLANATION OF ( E25 + E26 ) = 2356.3 eV = -E(3s2) ' Here the E(3s2) represents the binding energy of the two paired electrons (3s2) given by applying my formula of 2008. The charges (-10e) of the inner 10 electrons of 1s2.2s2.2p6 screen the nuclear charge (+36e) and for a perfect screening we would have an effective ζ = 26. However the two electrons of 3s2 penetrate the 2p6 leading to the deformation of shells with ζ > 26. Under this condition we write the following equation as ( E25 + E26 ) = 2356.3 eV = - E(3s2) = - 27.21)ζ2 + (16.95) ζ - 4.1 / n2 Since n = 3 we may write 3.0233ζ2 - 1.8833ζ - 2355.84 = 0 Then solving for ζ we get ζ = 28.23 > 26 . Note that the two electrons of opposite spin (4s2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy. However in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008. Category:Fundamental physics concepts